University of Wisconsin Madison Textbook Math 221

8/20/2019 University of Wisconsin Madison Textbook Math 221

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MATH 221FIRST SEMESTER

CALCULUS

Spring 2013

Typeset:January 17, 2013

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MATH 221 – 1st Semester CalculusLecture notes version 1.1 (Spring 2013)

Copyright (c) 2012 Sigurd B. Angenent, Laurentiu Maxim, Evan Dummit, Joel Robbin.Permission is granted to copy, distribute and/or modify this document under the termsof the GNU Free Documentation License, Version 1.2 or any later version published by

the Free Soware Foundation; with no Invariant Sections, no Front-Cover Texts, and noBack-Cover Texts. A copy of the license is included in the section entitled ”GNU Free

Documentation License”.


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Contents

Chapter I. Numbers and Functions 5

1. Numbers 5

2. Problems 9

3. Functions 10

4. Implicit functions 13

5. Inverse functions 16

6. Inverse trigonometric functions 17

7. Problems 20

Chapter II. Derivatives (1) 22

1. The tangent line to a curve 22

2. An example – tangent to a parabola 23

3. Instantaneous velocity 24

4. Rates of change 25

5. Examples of rates of change 25

6. Problems 27

Chapter III. Limits and Continuous

Functions 30

1. Informal definition of limits 30

2. Problems 32

3. The formal, authoritative, definition

of limit 324. Problems 36

5. Variations on the limit theme 37

6. Properties of the Limit 39

7. Examples of limit computations 40

8. When limits fail to exist 42

9. Limits that equal ∞ 4410. What’s in a name? – Free Variables

and Dummy variables 46

11. Limits and Inequalities 47

12. Continuity 48

13. Substitution in Limits 50

14. Problems 51

15. Two Limits in Trigonometry 52

16. Problems 54

17. Asymptotes 56

18. Problems 58

Chapter IV. Derivatives (2) 59

1. Derivatives Defined 59

2. Direct computation of derivatives 60

3. Differentiable implies Continuous 62

4. Some non-differentiable functions 62

5. Problems 63

6. The Differentiation Rules 64

7. Differentiating powers of functions 67

8. Problems 69

9. Higher Derivatives 70

10. Problems 71

11. Differentiating trigonometric

functions 72

12. Problems 73

13. The Chain Rule 74

14. Problems 80

15. Implicit differentiation 82

16. Problems 8517. Problems on Related Rates 85

Chapter V. Graph Sketching and Max-Min

Problems 89

1. Tangent and Normal lines to a graph 89

2. The Intermediate Value Theorem 89

3. Problems 91

4. Finding sign changes of a function 92

5. Increasing and decreasing functions 93

6. Examples 94

7. Maxima and Minima 99

8. Must a function always have a

maximum? 100

9. Examples – functions with and

without maxima or minima 10110. General method for sketching the

graph of a function 102

11. Convexity, Concavity and the

Second Derivative 104

12. Problems 106

13. Applied Optimization 108

14. Problems 109

15. Parametrized Curves 112

16. Problems 117

17. l’Hopital’s rule 118

18. Problems 120

Chapter VI. Exponentials and Logarithms

(naturally) 122

1. Exponents 122

2. Logarithms 124

3. Properties of logarithms 124

4. Graphs of exponential functions and

logarithms 125

5. The derivative of ax and the

definition of e 125

6. Derivatives of Logarithms 127

7. Limits involving exponentials and

logarithms 128

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4 CONTENTS

8. Exponential growth and decay 129

9. Problems 131

Chapter VII. The Integral 1351. Area under a Graph 135

2. When f changes its sign 137

3. The Fundamental Theorem of

Calculus 137

4. Summation notation 138

5. Problems 139

6. The indefinite integral 141

7. Properties of the Integral 142

8. The definite integral as a function of

its integration bounds 145

9. Substitution in Integrals 146

10. Problems 148

Chapter VIII. Applications of the integral 153

1. Areas between graphs 153

2. Problems 1553. Cavalieri’s principle and volumes of

solids 155

4. Three examples of volume

computations of solids of

revolution 160

5. Volumes by cylindrical shells 162

6. Problems 164

7. Distance from velocity 165

8. The arclength of a curve 168

9. Problems 170

10. Velocity from acceleration 170

11. Work done by a force 171

12. Problems 173

Answers and Hints 175


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CHAPTER I

Numbers and Functions

e subject of this course is “functions of one real variable” so we begin by talkingabout the real numbers, and then discuss functions.

1. Numbers

1.1. Different kinds of numbers. e simplest numbers are the positive integers

1, 2, 3, 4, · · ·the number zero

0,

and the negative integers

· · · , −4, −3, −2, −1.Together these form the integers or “whole numbers.”

Next, there are the numbers you get by dividing one whole number by another(nonzero) whole number. ese are the so called fractions or rational numbers , such as

1

2, 1

3,

2

3,

1

4,

2

4, 3

4, 4

3, · · ·

or

−1

2

,

−1

3

,

−2

3

,

−1

4

,

−2

4

,

−3

4

,

−4

3

,

· · ·By definition, any whole number is a rational number (in particular zero is a rationalnumber.)

You can add, subtract, multiply and divide any pair of rational numbers and the resultwill again be a rational number (provided you don’t try to divide by zero).

Approximating√

2 :

x x2

1.2 1 .44

1.3 1 .69

1.4 1.96 2

1.6 2 .56

ere are other numbers besides the rational numbers: it was discovered by the an-cient Greeks that the square root of 2 is not a rational number. In other words, there doesnot exist a fraction m

n such that(m

n

)2= 2, i.e. m2 = 2n2.

Nevertheless, if you compute x2 for some values of x between 1 and 2, and check if youget more or less than 2, then it looks like there should be some number x between 1.4and 1.5 whose square is exactly 2.

1.2. A reason to believe in√

2. e Pythagorean theorem says that the hypotenuseof a right triangle with sides 1 and 1 must be a line segment of length

√ 2. In middle or

high school you learned something similar to the following geometric construction of a

line segment whose length is√

2. Take a square with side of length 1, and construct a newsquare one of whose sides is the diagonal of the first square. e figure you get consistsof 5 triangles of equal area and by counting triangles you see that the larger square hasexactly twice the area of the smaller square. erefore the diagonal of the smaller square,

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6 I. NUMBERS AND FUNCTIONS

being the side of the larger square, is√

2 as long as the side of the smaller square.So, we will assume that there is such a number whose square is exactly 2, and we

call it the square root of 2, wrien as√

2. ere are more than a few questions¹ raisedby assuming the existence of numbers like the square root of 2, but we will not deal withthese questions here. Instead, we will take a more informal approach of thinking of realnumbers as “infinite decimal expansions”.

1.3. Decimals. One can represent certain fractions as finite decimal numbers, e.g.,

279

25 =

1116

100 = 11.16.

Some rational numbers cannot be expressed with a finite decimal expansion. For instance,expanding 13 as a decimal number leads to an unending “repeating decimal”:

1

3 = 0.333333333333333 · · ·It is impossible to write the complete decimal expansion of 1

3 because it contains infinitely

many digits. But we can describe the expansion: each digit is a 3.Every fraction can be wrien as a decimal number which may or may not be finite.

If the decimal expansion doesn’t terminate, then it will repeat (although this fact is notso easy to show). For instance,

1

7 = 0.142857 142857 142857 142857 . . .

Conversely, any infinite repeating decimal expansion represents a rational number.A real number is specified by a possibly unending decimal expansion. For instance,

√ 2 = 1.4142135623730950488016887242096980785696718753769. . .

Of course you can never write all the digits in the decimal expansion, so you only writethe first few digits and hide the others behind dots. To give a precise description of a real

number (such as√

2) you have to explain how you could in principle compute as manydigits in the expansion as you would like².

1.4. Why are real numbers called real? All the numbers we will use in this firstsemester of calculus are “real numbers”. At some point in history it became useful toassume that there is such a thing as

√ −1, a number whose square is −1. No real numberhas this property (since the square of any real number is nonnegative) so it was decidedto call this newly-imagined number “imaginary” and to refer to the numbers we already

had (rationals, √ 2-like things) as “real”.

¹Here are some questions: if we assume into existence a number a number x between 1.4 and 1.5 for

which x2 = 2 , how many other such numbers must we also assume into existence? How do we know there is“only one” such square root of 2? And how can we be sure that these new numbers will obey the same algebra

rules (like a + b = b + a) as the rational numbers?²What exactly we mean by specifying how to compute digits in decimal expansions of real numbers is

another issue that is beyond the scope of this course.


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1. NUMBERS 7

1.5. Reasons not to believe in ∞. In calculus we will oen want to talk about verylarge and very small quantities. We will use the symbol ∞ (pronounced “infinity”) all thetime, and the way this symbol is traditionally used would suggest that we are thinking of ∞ as just another number. But ∞ is different. e ordinary rules of algebra don’t applyto ∞. As an example of the many ways in which these rules can break down, just thinkabout “∞ + ∞.” What do you get if you add infinity to infinity? e elementary schoolargument for finding the sum is: “if you have a bag with infinitely many apples, and youadd infinitely many more apples, you still have a bag with infinitely many apples.” So,you would think that

∞ + ∞ = ∞.If ∞ were a number to which we could apply the rules of algebra, then we could cancel∞ from both sides,

∞+̸ ∞ ≠∞ =⇒ ∞ = 0.So infinity is the same as zero! If that doesn’t bother you, then let’s go on. Still assuming

∞is a number we find that

∞∞ = 1,but also, in view of our recent finding that ∞ = 0 ,

∞∞ =

0

∞ = 0.erefore, combining these last two equations,

1 = ∞∞ = 0.

In elementary school terms: “one apple is no apple.”is kind of arithmetic is not going to be very useful for scientists (or grocers), so we

need to drop the assumption that led to this nonsense, i.e. we have to agree from here onthat ³

INFINITY IS NOT A NUMBER!1.6. e real number line and intervals. It is customary to visualize the real num-

bers as points on a straight line. We imagine a line, and choose one point on this line,which we call the origin. We also decide which direction we call “le” and hence whichwe call “right.” Some draw the number line vertically and use the words “up” and “down.”

To plot any real number x one marks off a distance x from the origin, to the right(up) if x > 0, to the le (down) if x


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2. PROBLEMS 9

so that D is the set containing the numbers 1, 2, and 3. Or the set

E = {x | x3 − 4x2 + 1 = 0}which consists of the solutions of the equation x3 − 4x2 + 1 = 0. (ere are three of

them, but it is not easy to give a formula for the solutions.)If A andB are two sets then the unionof A and B is the set that contains all numbers

that belong either to A or to B. e following notation is used

A ∪ B = {x | x belongs to A or to B or both.}Similarly, the intersection of two sets A and B is the set of numbers that belong to bothsets. is notation is used:

A ∩B = {x | x belongs to both A and B.}2. Problems

1. What is the 2007th digit aer the period

in the expansion of 17? [ A]

2. Which of the following fractions have fi-nite decimal expansions?

a = 2

3, b =

3

25, c =

276937

15625 .

3. Draw the following sets of real numbers.Each ofthese sets is theunion of one or moreintervals. Find those intervals. Which of these sets are finite?

A =

x | x2 − 3x + 2 ≤ 0B = x | x

2 − 3x + 2 ≥ 0C =

x | x

2

− 3x > 3D =

x | x2 − 5 > 2x

E =

t | t2 − 3t + 2 ≤ 0F =

α | α2 − 3α + 2 ≥ 0

G = (0, 1) ∪ (5, 7]H =

({1} ∪ {2, 3} ∩ (0, 2√ 2)Q =

θ | sin θ = 1

2

R =

ϕ | cos ϕ > 0

4. Suppose A and B are intervals. Is it al-ways true that A ∩ B is an interval? Howabout A ∪B?5. Consider the sets

M =

x | x > 0 and N = y | y > 0.Are these sets the same? [ A]

6. [Group Problem] Write the numbers

x = 0.3131313131 . . . ,

y = 0.273273273273 . . .

and z = 0.21541541541541541 . . .

as fractions (that is, write them as mn

, spec-

ifying m and n.)

(Hint: show that 100x = x + 31 . A similartrick works for y, but z is a lile harder.) [ A]

7. [Group Problem] (a) In §1.5 we agreedthat infinitely large numbers don’t exist. Doinfinitely small numbers exist? In otherwords, does there exist a positive numberx that is smaller than 1

n for all n =

1, 2, 3, 4, · · · , i.e.0 < x < 1

2, and 0 < x < 1

3, and

0 < x < 14

, and so on

· · ·?

(b) Is the number whose decimal expansionaer the period consists only of nines, i.e.

a = 0.99999999999999999 . . .

the same as the number 1? Or could it bethat there are numbers between 0.9999 · · ·and 1? That is, is it possible that there issome number x that satisfies

0.999999 · · · < x


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3. Functions

3.1. Dependence. Calculus deals with quantities that change. For instance, the wa-

ter temperature T of Lake Mendota (as measured at the pier near the Memorial Union)is a well-defined quantity, but it changes with time. At each different time t we will finda different temperature T . erefore, when we say “the temperature at the pier of LakeMendota,” we could mean two different things:

• On one hand we could mean the “temperature at some given time,” e.g. the tem-perature at 3pm is 68F: here the temperature is just a number. e most commonnotation for this is T (3) = 68, or T (3pm) = 68F.

• On the other hand we could mean the “temperature in general,” i.e. the temper-atures at all times. In that second interpretation the temperature is not just anumber, but a whole collection of numbers, listing all times t and the corre-sponding temperatures T (t).

So T (t) is a number while T by itself is not a number, but a more complicated thing.

It is what in mathematics is called a function. We say that the water temperature is a function of time.

Here is the definition of what a mathematical function is:

3.2. Definition. To specify a function f you must

(1) give a rule that tells you how to compute the value f (x) of the function for a givenreal number x, and:

(2) say for which real numbers x the rule may be applied.

e set of numbers for which a function is defined is called its domain. e set of all possible numbers f (x) as x runs over the domain is called the range of the function. e rule must be unambiguous: the same x must always lead to the same f (x).

For instance, one can define a function f by puing f (x) = 3x for all x

≥ 0. Here

the rule defining f is “multiply by 3 whatever number you’re given”, and the function f will accept all real numbers.

e rule that specifies a function can come in many different forms. Most oen it isa formula, as in the square root example of the previous paragraph. Sometimes you needa few formulas, as in

g(x) =

2x for x


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3. FUNCTIONS 11

range of f

y = f (x) (x, f (x))

x

domain of f

Figure 3. The graph of a function f . The domain of f consists of all x values at which the functionis defined, and the range consists of all possible values f can have.

P 0

P 1

y1 − y0

x1 − x0

x0

x1

y0

y1

n

1

m

Figure 4. The graph of f (x) = mx + n is a straight line. It intersects the y-axis at height n. Theratio between the amounts by which y and x increase as you move from one point to another onthe line is y1−y0

x1−x0 = m. This ratio is the same, no maer how you choose the points P 0 and P 1 aslong as they are different and on the line.

function. If we know two points (x0, y0) and (x1, y1) on the line, then we can computethe slope m from the “rise-over-run” formula

m = y1 − y0x1 − x0 .

is formula actually contains a theorem from Euclidean geometry, namely, it says thatthe ratio

(y1 − y0) : (x1 − x0)is the same for every pair of distinct points (x0, y0) and (x1, y1) that you could pick onthe line.

3.5. Domain and “biggest possible domain.” In this course we will usually not becareful about specifying the domain of a function. When this happens the domain is


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understood to be the set of all x for which the rule that tells you how to compute f (x) isa meaningful real number. For instance, if we say that h is the function

h(x) = √ xthen the domain of h is understood to be the set of all nonnegative real numbers

domain of h = [0, ∞)since

√ x is a well-defined real number for all x ≥ 0 and not a real number for x 0 then y = 1/x2 has a solution (in fact two solutions), namely x = ±1/√ y. isshows that the range of f is

“all positive real numbers” = {x | x > 0} = (0, ∞).3.7. Functions in “real life”. One can describe the motion of an object using a func-

tion. If some object is moving along a straight line, then you can define the following

function: Let s(t) be the distance from the object to a fixed marker on the line, at the timet. Here the domain of the function is the set of all times t for which we know the positionof the object, and the rule is

Given t, measure the distance between the object at time t and the marker.

ere are many examples of this kind. For instance, a biologist could describe the growthof a mouse by defining m(t) to be the mass of the mouse at time t (measured since the

birth of the mouse). Here the domain is the interval [0, T ], where T is the lifespan of themouse, and the rule that describes the function is

Given t, weigh the mouse at time t.h

Here is another example: suppose you are given an hourglass. If you turn it over, thensand will pour from the top part to the boom part. At any time t you could measurethe height of the sand in the boom and call it h(t). en, as in the previous examples,you can say that the height of the sand is a function of time. But in this example youcan let the two variables height and time switch roles: given a value for h you wait until


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4. IMPLICIT FUNCTIONS 13

y = x3 − x

Figure 5. The graph of y = x3 − x fails the “horizontal line test,” but it passes the “vertical linetest.” The circle fails both tests.

the pile of sand in the boom has reached height h and check what time it is when thathappens: the resulting time t(h) is determined by the specified height h. In this way youcan regard time as a function of height.

3.8. e Vertical Line Property. Generally speaking, graphs of functions are curvesin the plane but they distinguish themselves from arbitrary curves by the way they in-tersect vertical lines: e graph of a function cannot intersect a vertical line “ x =constant” in more than one point . e reason why this is true is very simple: if twopoints lie on a vertical line, then they have the same x coordinate, so if they also lie onthe graph of a function f , then their y-coordinates must both be equal to f (x), so in factthey are the same point.

3.9. Example – a cubic function. e graph of f (x) = x3 − x “goes up and down,”and, even though it intersects several horizontal lines in more than one point, it intersectsevery vertical line in exactly one point. See Figure 5.

3.10. Example – a circle is not a graph of a function y = f (x). e collection of points determined by the equation x2 + y2 = 1 is a circle. It is not the graph of a functionsince the vertical line x = 0 (the y-axis) intersects the graph in two points P 1(0, 1) andP 2(0, −1). See again Figure 5. is example continues in § 4.3 below.

4. Implicit functions

For many functions the rule that tellsyou how to compute it is not an explicit formula,but instead an equation that you still must solve. A function that is defined in this way iscalled an “implicit function.”

4.1. Example. We can define a function f by saying that if x is any given number,then y = f (x) is the solution of the equation

x2 + 2y − 3 = 0.In this example we can solve the equation for y,

y = 3 − x2

2 .

us we see that the function we have defined is f (x) = (3 − x2)/2.Here we have two definitions of the same function, namely

(i) “y = f (x) is defined by x2 + 2y − 3 = 0,” and(ii) “f is defined by f (x) = (3 − x2)/2.”


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e first definition is an implicit definition, the second is explicit. is example showsthat with an “implicit function” it is not the function itself, but rather the way it wasdefined that is implicit.

4.2. Another example: domain of an implicitly defined function. Define g by say-ing that for any x the value y = g(x) is the solution of

x2 + xy − 3 = 0. Just as in the previous example you can then solve for y, and you find that

g(x) = y = 3 − x2

x .

Unlike the previous example this formula does not make sense when x = 0, and indeed,for x = 0 our rule for g says that g(0) = y is the solution of

02 + 0 · y − 3 = 0, i.e. y is the solution of 3 = 0.at equation has no solution and hence x

= 0 does not belong to the domain of our

function g.

x2 + y2 = 1y = +

√ 1 − x2

y = −√ 1 − x2

Figure 6. The circle determined by x2 + y2 = 1 is not the graph of a function, but it contains thegraphs of the two functions h1(x) =

√ 1 − x2 and h2(x) = −

√ 1 − x2.

4.3. Example: the equation alone does not determine the function. We saw in§ 3.10 that the unit circle is not the graph of a function (because it fails the vertical linetest). What happens if you ignore this fact and try to use the equation x2 + y2 = 1 forthe circle to define a function anyway? To find out, suppose we define y = h(x) to be“the solution” of

x2 + y2 = 1.

If x > 1 or x 1 and there is no solution, so h(x) is at most definedwhen

−1

≤ x

≤ 1. But when

−1 < x


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4. IMPLICIT FUNCTIONS 15

To see this in the picture, look at Figure 6 and find all vertical lines that intersect the circleon the le exactly once.

To get different functions that are described by the equation x2 + y2 = 1, we haveto specify for each x which of the two solutions ±√ 1 − x2 we declare to be “f (x)”. isleads to many possible choices. Here are three of them:

h1(x) = the non negative solution y of x2 + y2 = 1

h2(x) = the non positive solution y of x2 + y2 = 1

h3(x) =

h1(x) when x


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Don’t worry about how this formula came about; let’s just trust Cardano and Tartaglia.e implicit description (2) looks a lot simpler, and when we try to differentiate thisfunction later on, it will be much easier to use “implicit differentiation” than to use theCardano-Tartaglia formula directly.

Finally, you could have been given the function h whose definition is

y = h(x) if and only if sin(y) + 3y + 2x = 0. (3)

ere is no formula involving only standard functions (exponents, trig and inverse trigfunctions, logarithms, etc. for the solution to this equation. Nonetheless it turns out thatno maer how you choose x, the equation sin(y) + 3y + 2x = 0 has exactly one solutiony; in fact, you will prove this in Problem 54. So the function h is well defined, but for thisfunction the implicit description is the only one available.

5. Inverse functions

If we have a function f , which takes input values and sends them to an output, we

might want to try to define a function f −1 which “undoes” f , by the following prescrip-tion:

For any given x we say that y = f −1(x)if y is the solution of f (y) = x.

(4)

Note that x and y have swapped their usual places in this last equation!e prescription (4) defines the inverse function f −1, but it does not say what the

domain of f −1 is. By definition, the domain of f −1 consists of all numbers x for which the equation f (y) = x has exactly one solution. ‘ So if for some x the equation f (y) = x hasno solution y, then that value of x does not belong to the domain of f −1.

If, on the other hand, for some x the equation f (y) = x has more than one solutiony, then the prescription (4) for computing f −1(x) is ambiguous: which of the solutionsy should be f −1(x)? When this happens we throw away the whole idea of finding theinverse of the function f , and we say that the inverse function f −1 is undefined (“thefunction f has no inverse”.)

5.1. Example – inverse of a linear function. Consider the function f with f (x) =2x + 3. en the equation f (y) = x works out to be

2y + 3 = x

a

f (a)

f (a)

a

b

f (b)

f (b)

b

c

f (c)

f (c)

c

The graph of f

The graph of f −1

Figure 7. The graph of a function and its inverse are mirror images of each other. Can you drawthe mirror?


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6. INVERSE TRIGONOMETRIC FUNCTIONS 17

and this has the solution

y = x − 3

2 .

So f −1(x) is defined for all x, and it is given by f −1(x) = (x − 3)/2.5.2. Example – inverse of f (x) = x2. It is oen said that “the inverse of x2 is

√ x.”

is is not quite true, as you’ll see in this and the next example.Let f be the function f (x) = x2 with domain all real numbers. What is f −1?e equation f (y) = x is in this case y2 = x. When x > 0 the equation has two

solutions, namely y = +√

x and y = −√ x. According to our definition, the function f does not have an inverse.

5.3. Example – inverse of x2, again. Consider the function g(x) = x2 with domainall positive real numbers. To see for which x the inverse g−1(x) is defined we try to solvethe equation g(y) = x, i.e. we try to solve y2 = x. If x < 0 then this equation has nosolutions since y2

≥ 0 for all f . But if x

≥ 0 then y2 = x does have a solution, namely

y = √ x.So we see that g−1(x) is defined for all positive real numbers x, and that it is given

by g−1(x) =√

x.is example is shown in Figure 7. See also Problem 6.

6. Inverse trigonometric functions

e two most important inverse trigonometric functions are the arcsine and thearctangent . e most direct definition of these functions is given in Figure 8. In words,θ = arcsin x is the angle (in radians) whose sine is x. If −1 ≤ x ≤ 1 then there always issuch an angle, and, in fact, there are many such angles. To make the definition of arcsin x

1x

arcsin x

x

arctan x

1

+1

−1

−∞

+

∞

arcsin x is defined

for any x from −1to +1

arctan x is defined

for any x from

−∞ to +∞

e arc whose sine is x

e arc whose tangent is x

Figure 8. Definition of arcsin x and arctan x. The doed circles are unit circles. On the le asegment of length x and its arc sine are drawn. The length of the arc drawn on the unit circle isthe subtended angle in radians, i.e. arcsin x. So arcsin x is the length of “the arc whose sine is x.”


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π/2−π/21

−1 y = sin x

π/2

−π/2

1

−1

y = arcsin x

− 3π2 −π2 π2 3π2

y = tan x

−π/2

π/2

y = arctan x

Figure 9. The graphs of the sine and tangent functions on the le, and their inverses, the arcsineand arctangent on the right. Note that the graph of arcsine is a mirror image of the graph of thesine, and that the graph of arctangent is a mirror image of the graph of the tangent.

unambiguous we always choose θ to be the angle that lies between −π2 and + π2 . To seewhere the name “arcsine” comes from, look at Figure 8 on the le.

An equivalent way of defining the arcsine and arctangent is to say that they arethe inverse functions of the sine and tangent functions on a restricted domain. E.g. if y = f (x) = sin x, then the inverse of the function f is by definition (see (4)) the functionf −1 with the property that

y = f −1(x) ⇐⇒ x = f (y) = sin y.If we restrict y to the interval −π2 ≤ y ≤ π2 then this is just the definition of arcsin x, so

y = sin x ⇐⇒ x = arcsin y, provided − π2 ≤ x ≤ π2 .Likewise,

y = tan x ⇐⇒ x = arctan y, provided − π2 ≤ x ≤ π2 .Forgeing about the requirement that −π2 ≤ x ≤ π2 can lead to unexpected mistakes (seeProblem 8).

Because of the interpretation of y = arcsin x as the inverse of the sine function, thenotations

arcsin x = sin−1 x, arctan x = tan−1 x

are very commonly used.


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6. INVERSE TRIGONOMETRIC FUNCTIONS 19

In addition to the arcsine and arctangent, people have also defined the arccosine, thearcsecant and the arccosecant. However, because they can all be expressed in terms of the arcsine and arctangent, we will not bother with them.


20/195


7. Problems

1. The functions f and g are defined by

f (x) = x2 and g(s) = s2.

Are f and g the same functions or are theydifferent? [ A]

2. Find a formula for the function f that isdefined by

y = f (x) ⇐⇒ x2y + y = 7.What is the domain of f ?

3. Find a formula for the function f that isdefined by the requirement that for any xone has

y = f (x) ⇐⇒ x2y − y = 6.What is the domain of f ?

4. Let f be the function defined by the re-quirement that for any x one has

y = f (x) ⇐⇒y is the largest of allpossible solutions of y2 = 3x2 − 2xy.

Find a formula for f . What are the domainand range of f ? [ A]

5. Find a formula for the function f that isdefined by

y = f (x) ⇐⇒ 2x + 2xy + y2 = 5and y > −x.

Find the domain of f .

6. (continuation of example 6.) Let k be thefunction with k(x) = x2 whose domain isall negative real numbers. Find the domainof k−1, and draw the graph of k−1. [ A]

7. Use a calculator to compute g(1.2) inthree decimals where g is the implicitly de-fined function from §4.4. (There are (at least)two different ways of finding g(1.2))

8. [Group Problem] True or false?(a) For all real numbers x one has

sin(

arcsin x

= x.

[ A]

(b) For all real numbers x one has

arcsin(

sin x

= x.

[ A]

(c) For all real numbers x one has

arctan(

tan x

= x.

[ A]

(d) For all real numbers x one has

tan(

arctan x

= x.

[ A]

9. On a graphing calculator plot the graphsof the following functions, and explain the

results. (Hint: first do the previous exercise.)

f (x) = arcsin(sin x), −2π ≤ x ≤ 2πg(x) = arcsin(x) + arccos(x), 0 ≤ x ≤ 1h(x) = arctan

sin x

cos x, |x| < π /2

k(x) = arctan cos x

sin x, |x| < π /2

l(x) = arcsin(cos x), −π ≤ x ≤ πm(x) = cos(arcsin x), −1 ≤ x ≤ 1

10. Find the inverse of the function f that isgiven by f (x) = sin x and whose domain is π

≤ x

≤ 2π. Sketch the graphs of both f

and f −1.

11. Find a number a such that the functionf (x) = sin(x + π/4) with domain a ≤ x ≤a + π has an inverse. Give a formula forf −1(x) using the arcsine function.

12. Simplicio has found a new formula for thearcsine. His reasoning is as follows:

Since everybody writes “the square of sin y” as (sin y

2= sin2 y.

we can replace the 2’s by −1’s and we get arcsin y = sin−1 y = (sin y−1 = 1

sin y.

Is Simplicio right or wrong? Explain youropinion.

13. Draw the graph of the function h3 from§4.3.

14. A function f is given that satisfies

f (2x + 3) = x2


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7. PROBLEMS 21

for all real numbers x.

If x and y are arbitrary real numbers then

compute(a) f (0) [ A]

(b) f (3) [ A]

(c) f (π) [ A]

(d) f (t) [ A]

(e) f (x) [ A]

(f ) f (f (2)) [ A]

(g) f (2f (x)) [ A]

15. A function f is given that satisfies

f ( 1

x + 1

= 2x − 12

for all real numbers x.

If x and t are arbitrary real numbers, thencompute the following quantities:

(a) f (1) [ A]

(b) f (0) [ A]

(c) f (t) [ A]

(d) f (x) [ A]

(e) f (f (2)) [ A]

(f ) f (2f (x)) [ A]

16. Does there exist a function f that satis-fies

f (x2) = x + 1

for all real numbers x? [ A]

∗ ∗ ∗

The followingexercises review precalculusma- terial involving quadratic expressions ax2 +bx + c in one way or another.

17. Find the vertex (h, k) of the parabolay = ax2 + bx + c. Use the result to findthe range of this function. Note that the be-havior depends on whether a is positive ornegative. (Hint: “Complete the square” inthe quadratic expression by writing it in theform y = a(x − h)2 + k for some h and kin terms of a, b, and c.)

18. Find the ranges of the following func-tions:

f (x) = 2x2 + 3

g(x) = −2x2

+ 4xh(x) = 4x + x2

k(x) = 4 sin x + sin2 x

ℓ(x) = 1/(1 + x2)

m(x) = 1/(3 + 2x + x2).

[ A]

19. [Group Problem] For each real num-ber a we define a line ℓa with equation y =ax + a2.

(a) Draw the lines corresponding to a =

−2, −1, −12 , 0, 12 , 1, 2.(b) Does thepoint with coordinates (3, 2) lieon one or more of the lines ℓa (where a canbe any number, not just the five values frompart (a))? If so, for which values of a does

(3, 2) lie on ℓa?

(c) Which points in the plane lie on at leastone of the lines ℓa?.

20. For which values of m and n does thegraph of f (x) = mx +n intersect the graphof g(x) = 1/x in exactly one point and alsocontain the point (−1, 1)?

21. For which values of m and n does thegraph of f (x) = mx + n not intersect thegraph of g(x) = 1/x?


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CHAPTER II

Derivatives (1)

To work with derivatives we first have to know what a limit is, but to motivate whywe are going to study limits we essentially need to explain what a derivative is. In orderto resolve this potentially circular logic, we will first motivate the idea of a derivative inthis short chapter, then discuss how to compute limits in detail in the next chapter, andthen we will return and discuss derivatives armed with a beer understanding of limits.

Let’s first look at the two classical problems that gaveriseto the notion of a derivative:

finding the equation of the line tangent to a curve at a point, and finding the instantaneousvelocity of a moving object.

1. e tangent line to a curve

Suppose you have a function y = f (x) and you draw its graph. If you want to findthe tangent line to the graph of f at some given point on the graph of f , how would youdo that?

P

Q

tangent

a secant

Figure 1. Constructing the tangent by leing Q → P

22


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2. AN EXAMPLE – TANGENT TO A PARABOLA 23

Let P be the point on the graph at which want to draw the tangent. If you are makinga real paper and ink drawing you would take a ruler, make sure it goes through P , andthen turn it until it doesn’t cross the graph anywhere else (at least, nowhere else near P ).

If you are using equations to describe the curve and lines, then you could pick apoint Q on the graph and construct the line through P and Q (“construct” means “findan equation for”). is line is called a “secant line”, and it won’t precisely be the tangentline, but if you choose Q to be very close to P then the secant line will be close to thetangent line.

So this is our recipe for constructing the tangent through P : pick another point Q onthe graph, find the line through P and Q, and see what happens to this line as you takeQ closer and closer to P . e resulting secants will then get closer and closer to someline, and that line is the tangent.

We’ll write this in formulas in a moment, but first let’s worry about how close Qshould be to P . We can’t set Q equal to P , because then P and Q don’t determine aline, since we need two points to determine a line. If we choose Q different from P then

we won’t get the tangent, but at best something that is “close” to it. Some people havesuggested that one should take Q “infinitely close” to P , but it isn’t clear what that wouldmean. e concept of a limit is needed to clarify this issue.

2. An example – tangent to a parabola

To make things more concrete, let us take the function f (x) = x2, and aempt tofind the equation of the tangent line to y = f (x) at the point P = (1, 1). e graph of f is of course a parabola.

Any line through the point P (1, 1) has equation

y − 1 = m(x − 1)where m is the slope of the line. So instead of finding the equations of the secant andtangent lines, we can simply find their slopes.

1

1

x

x2

Δx

Δy = x2-1

y = x 2

Q

P

To find the slope of the tangent at P(1,1)

pick another point Q (x, x2) on the parabola

and compute the slope mPQ of the line

connecting P and Q.

Then let Q approach P and see what

happens to the slope mPQ .

Finding the slope of the tangent at P

Let Q be the other point on the parabola, with coordinates (x, x2). We can “move Qaround on the graph” by changing x, although we are not allowed to set x = 1 because P

and Q have to be different points. By the “rise over run” formula, the slope of the secantline joining P and Q is

mPQ = ∆y

∆x where ∆y = x2 − 1 and ∆x = x − 1.

By factoring x2 − 1 we can rewrite the formula for the slope as follows

mPQ = ∆y

∆x =

x2 − 1x − 1 =

(x − 1)(x + 1)x − 1 = x + 1. (5)


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24 II. DERIVATIVES (1)

As x gets closer to 1, the slope mPQ, being x + 1, gets closer to the value 1 + 1 = 2. Wesay that

the limit of the slope mPQ

as Q approaches P is 2.

In symbols,limQ→P

mPQ = 2,

or, since Q approaching P is the same as x approaching 1,

limx→1

mPQ = 2. (6)

So we find that the tangent line to the parabola y = x2 at the point (1, 1) has equation

y − 1 = 2(x − 1), i.e. y = 2x − 1.A warning: we cannot substitute x = 1 in equation (5) to get (6), even though it lookslike that’s what we did. e reason why we can’t do that is that when x = 1 the pointQ coincides with the point P so “the line through P and Q” is not defined; also, if x = 1then ∆x = ∆y = 0 so that the rise-over-run formula for the slope gives

mPQ = ∆y

∆x =

0

0 = undefined.

It is only aer the algebra trick in (5) that seing x = 1 gives something that is well-defined. But if the intermediate steps leading to mPQ = x + 1 aren’t valid for x = 1,why should the final result mean anything for x = 1?

We did something more complicated than just seing x = 1: we did a calculationwhich is valid for all x ̸= 1, and later looked at what happens if x gets “very close to 1.”is is the essence of a limit, and we’ll study these ideas in detail soon.

3. Instantaneous velocity

When you are riding in a car, the speedometer tells you how fast you are going, i.e.what your velocity is. But what exactly does it mean when the speedometer says your

car is traveling at a speed of (say) 50 miles per hour?

We all know what average velocity is. Namely, if it takes you two hours to cover100 miles, then your average velocity was

distance traveled

time it took = 50 miles per hour.

is is not the number the speedometer provides you – it doesn’t wait two hours, measurehow far you went, and then compute distance/time. If the speedometer in your car tellsyou that you are driving 50mph, then that should be your velocity at the moment thatyou look at your speedometer, i.e. “distance traveled over time it took” at the momentyou look at the speedometer. But during the moment you look at your speedometer notime goes by (because a moment has no length) and you didn’t cover any distance, soyour velocity at that moment is 0

0, which is undefined. Your velocity at any moment is

undefined. But then what is the speedometer telling you?


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5. EXAMPLES OF RATES OF CHANGE 25

To put all this into formulas we need to introduce some notation. Let t be the time(in hours) that has passed since we got onto the road, and let s(t) be our distance fromour starting point (in miles).

Instead of trying to find the velocity exactly at time t, we find a formula for theaverage velocity during some (short) time interval beginning at time t. We’ll write ∆t forthe length of the time interval.

At time t we are s(t) miles from our start. A lile later, at time t+∆t we are s(t+∆t)miles from our start. erefore, during the time interval from t to t + ∆t, we have moved

s(t + ∆t) − s(t) miles,and therefore our average velocity in that time interval was

s(t + ∆t) − s(t)∆t

miles per hour.

e shorter we make the time interval (the smaller we choose ∆t) the closer this numbershould be to the instantaneous velocity at time t.

So we have the following formula (definition, really) for the velocity at time t

v(t) = lim∆t→0

s(t + ∆t) − s(t)∆t

. (7)

4. Rates of ange

e two previous examples have much in common. If we ignore all the details aboutgeometry, graphs, highways and motion, the following happened in both examples:

We had a function y = f (x), and we wanted to know how much f (x) changes if xchanges. If we change x to x + ∆x, then y will change from f (x) to f (x + ∆x). echange in y is therefore

∆y = f (x + ∆x) − f (x),and the average rate of change is

∆y

∆x =

f (x + ∆x) − f (x)∆x

. (8)

is is the average rate of change of f over the interval from x to x + ∆x. To define therate of ange of the function f at x we let the length ∆x of the interval become smallerand smaller, in the hope that the average rate of change over the shorter and shorter timeintervals will get closer and closer to some number. If that happens then that “limitingnumber” is called the rate of change of f at x, or, the derivative of f at x. It is wrien as

f ′(x) = lim∆x→0

f (x + ∆x) − f (x)∆x

. (9)

Derivatives and what we can do with them are what the first portion of this course isabout. e description we just went through shows that to understand what a deriva-

tive is, we need to understand more about this “limiting process” so that we can have aconcrete understanding of statements like (9).

5. Examples of rates of ange

5.1. Acceleration as the rate at whi velocity anges. As you are driving in yourcar your velocity may change over time. Suppose v(t) is your velocity at time t (measuredin miles per hour). You could try to figure out how fast your velocity is changing bymeasuring it at one moment in time (you get v(t)), then measuring it a lile later (you


26/195


get v(t + ∆t))). You conclude that your velocity increased by ∆v = v(t + ∆t) − v(t)during a time interval of length ∆t, and hence

average rate atwhich your

velocity changed

=

change in velocity

duration of time interval =

∆v

∆t =

v(t + ∆t) − v(t)∆t

.

is rate of change is called your average acceleration (over the time interval from t tot + ∆t). Your instantaneous acceleration at time t is the limit of your average accelerationas you make the time interval shorter and shorter:

{acceleration at time t} = a = lim∆t→0

v(t + ∆t) − v(t)∆t

.

e average and instantaneous accelerations are measured in “miles per hour per hour”:

(mi/h)/h = mi/h2.

Or, if you had measured distances in meters and time in seconds then velocities would bemeasured in meters per second , and acceleration in meters per second per second , which isthe same as meters per second2: “meters per squared second”.

5.2. Reaction rates. Imagine a chemical reaction in which two substances A and Breact in such a way that A converts B into A. e reaction could proceed by

A + B −→ 2A.If the reaction is taking place in a closed reactor, then the “amounts” of A and B willchange with time. e amount of B will decrease, while the amount of A will increase.Chemists write [A] for the concentration of “A” in the chemical reactor (measured inmoles per liter). We’re mathematicians so we will write “[A](t)” for the concentration of A present at time t.

A

B

A A

AB

Figure 2. A chemical reaction in which A converts B into A.

To describe how fast the amount of A is changing we consider the derivative of [A]with respect to time:

[A]′(t) = lim∆t→0

[A](t + ∆t) − [A](t)∆t

.

is quantity is the rate of change of [A]. In chemistry and physics, it is more commonto write the derivative in L notation:

d[A]

dt .


27/195

6. PROBLEMS 27

How fast does the reaction take place? If you add more A or more B to the reactor thenyou would expect that the reaction would go faster (i.e., that more A would be producedper second). e law of mass-action kinetics from chemistry states this more precisely.For our particular reaction it would say that the rate at which A is consumed is given by

d[A]

dt = k [A] [B],

where the constant k is called the reaction constant , which you could measure by timinghow fast the reaction goes.

6. Problems

1. Repeat the reasoning in §2 to find theslope of the tangent line at the point ( 1

3, 19

),

or more generally at any point (a, a2) on theparabola with equation y = x2.

2. Repeat the reasoning in §2 to find theslope of the tangent line at the point ( 1

2, 18

),

or more generally at any point (a, a3) on thecurve with equation y = x3.

3. Simplify the algebraic expressions you get

when you compute ∆y and ∆y/∆x for thefollowing functions

(a) y = x2 − 2x + 1(b) y =

1

x(c) y = 2x

[ A]

4. This figure shows a plot of the distancetraveled s(t) (in miles) versus time t (in min-utes):

s (t )

t

10 20 30 40 50

30

60

90

120

150

(a) Something is wrong: the curve in thegraph obviously doesn’t pass the verticalline test, so it cannot be the graph of a func-tion. How can it be the graph of s(t) versust? [ A]

(b) Use the plot to estimate the instanta-neous velocity at the following times

t (min) v(t)

30

60

90120

Describe in one or two short sentences whatyou did to find your estimates. [ A]

(c) Make a graphof the instantaneous veloc-ity v(t).


28/195


5. Look ahead at Figure 3 in the next chap-ter. What is the derivative of f (x) =x cos π

x at the points A and B on the graph?

[ A]

6. Suppose that some quantity y is a func-tion of some other quantity x, and supposethat y is a mass (measured in pounds) and

x is a length (measured in feet). What unitsdo the increments ∆y and ∆x, and the de-rivative dy/dx, have? [ A]

7. A tank is filling with water. The volume(in gallons) of water in the tank at time t(seconds) is V (t). What units does the de-rivative V ′(t) have? [ A]

8. [Group Problem] Let A(x) be the areaof an equilateral triangle whose sides mea-sure x inches.

(a) Show that dAdx

has the units of a length.

(b) Which length does dAdx

represent geomet-rically? [Hint: draw two equilateral trian-gles, one with side x and another with sidex + ∆x. Arrange the triangles so that theyboth have the origin as their lower le handcorner, and so their bases are on the x-axis.]

[ A]

9. [Group Problem] Let A(x) be the areaof a square with side x, and let L(x) be theperimeter of the square (sum of the lengthsof all its sides). Using the familiar formu-las for A(x) and L(x) show that A′(x) =12

L(x).

Give a geometric interpretation that ex-plains why ∆A ≈ 1

2L(x)∆x for small ∆x.

10. Let A(r) be the area enclosed by a circleof radius r , and let L(r) be the circumfer-ence of the circle. Show that A′(r) = L(r).

(Use the familiar formulas from geometryfor the area and circumference of a circle.)

11. Let V (r) be the volume enclosed by asphere of radius r, and let S (r) be the itssurface area.

(a) Show that V ′(r) = S (r). (Use the for-mulas V (r) = 4

3πr3 and S (r) = 4πr2.)

(b) Give a geometric explanation of the factthat dV

dr = S .

[Hint: to visualize what happens to the vol-ume of a sphere when you increase the ra-dius by a very small amount, imagine thesphere is the Earth, and you increase the ra-dius by covering the Earth with a layer of water that is 1 inch deep. How much doesthe volume increase? What if the depth of the layer was “∆r”?]

12. [Group Problem] Should you trust your calculator?

Find the slope of the tangent to the parabola

y = x2

at the point ( 13

, 19

) (You have already donethis: see exercise 1).

Instead of doing the algebra you could tryto compute the slope by using a calculator.This exercise is about how you do that and

what happens if you try (too hard).

Compute ∆y∆x

for various values of ∆x:

∆x = 0.1, 0.01, 0.001, 10−6, 10−12.

As you choose ∆x smaller your computed∆y∆x

ought to get closer to the actual slope.Use at least 10 decimals and organize yourresults in a table like this: Look carefully atthe ratios ∆y/∆x. Do they look like theyare converging to some number? Compare

the values of ∆y∆x

with the true value you gotin the beginning of this problem.


29/195

6. PROBLEMS 29

∆x a + ∆x f (a + ∆x) ∆y ∆y/∆x

0.1

0.01

0.001

10-6

10-12

Table 1. The table for Problem 12. Approximate the derivative of f (x) = x2 at a = 1/3 ≈0.333333333333 by computing

(f (a + ∆x) − f (a)/∆x for smaller and smaller values of ∆x.

You know from Problem 1 that the derivative is f ′(a) = 2a = 2/3. Which value of ∆x above givesyou the most accurate answer?


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CHAPTER III

Limits and Continuous Functions

While it is easy to define precisely in a few words what a square root is (√

a is thepositive number whose square is a) the definition of the limit of a function is more com-plicated. In this chapter we will take three approaches to defining the limit.

e first definition (§1) appeals to intuition and puts in words how most people thinkabout limits. Unfortunately this definition contains language that is ambiguous and themore you think about it the more you realize that it actually doesn’t mean anything. It is

also not clear enough to answer all questions that come up about limits.Most of the calculus you’ll see in this semester was essentially invented in the 17th

century, but the absence of a good definition of what derivatives and limits are led to cen-turies of confused arguments between experts. ese more or less ended when a precisedefinition was developed in the late 19th century, 200 years aer calculus was born! In §3you’ll see this precise definition. It runs over several terse lines, and unfortunately mostpeople don’t find it very enlightening when they first see it.

e third approach we will take is the axiomatic approa : instead of worryingabout the details of what a limit is, we use our intuition and in § 6 we write down anumber of properties that we believe the limit should have. Aer that we try to base allour reasoning on those properties.

1. Informal definition of limits

1.1. Definition of limit (1st attempt). If f is some function then

limx→a

f (x) = L

is read “the limit of f (x) as x approaches a is L.” It means that if you choose values of xthat are close but not equal to a, then f (x) will be close to the value L; moreover, f (x) gets closer and closer to L as x gets closer and closer to a.

e following alternative notation is sometimes used

f (x) → L as x → a;(read “f (x) approaches L as x approaches a” or “f (x) goes to L as x goes to a”.)

Note that in the definition we require x to approach a without ever becoming equal toa. It’s important that x never actually equals a because our main motivation for lookingat limits was the definition of the derivative. In Chapter II, equation (9) we defined thederivative of a function as a limit in which some number ∆x goes to zero:

f ′(x) = lim∆x→0

f (x + ∆x) − f (x)∆x

.

e quantity whose limit we want to take here is not even defined when ∆x = 0. ere-fore any definition of limit we come up with had beer not depend on what happens at∆x = 0; likewise, the limit limx→a f (x) should not depend on what f (x) does at x = a.

30


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1. INFORMAL DEFINITION OF LIMITS 31

1.2. Example. If f (x) = x + 3 then

limx

→4

f (x) = 7,

is “true”, because if you substitute numbers x close to 4 in f (x) = x + 3 the result willbe close to 7.

1.3. A complaint. Our first definition relies heavily on the phrase “gets close to” or“gets closer and closer to”. What does this mean? When x gets closer and closer to awithout ever being equal to a, how long does this take? (“are we there yet?”) How closeis close enough? Is it enough for x and a to be the same in five decimals? Fiy decimals?It is hard to answer these questions without generating new ones.

If we want to deal with limits with some measure of confidence that what we are do-ing isn’t ultimately nonsense, then we will need a beer definition of limit. Before goinginto that, let’s look at a practical approach to finding limits. To compute limx→a f (x) weneed to let x get closer and closer to a; we don’t really know how to do that, but we cancertainly grab a calculator or a computer and compute f (x) for several values of x that

are somewhat close to a (a to two decimals, a to three decimals, etc.) If the values of f (x)then begin to look like some fixed number we could guess that that number is the limit ¹.

Here are two examples where we try to find a limit by calculating f (x) for a fewvalues of x:

1.4. Example: substituting numbers to guess a limit. What (if anything) is

limx→2

x2 − 2xx2 − 4 ?

Here f (x) = (x2 − 2x)/(x2 − 4) and a = 2.We first try to substitute x = 2, but this leads to

f (2) = 22 − 2 · 2

22

−4

= 0

0

which does not exist. Next we try to substitute values of x close but not equal to 2. Table1 suggests that f (x) approaches 0.5.

x f (x)

3.000000 0.6000002.500000 0.5555562.100000 0.5121952.010000 0.5012472.001000 0.500125↓ ↓2 limit?

x g(x)

1.000000 1.0099900.500000 1.0099800.100000 1.0098990.010000 1.0089910.001000 1.000000↓ ↓0 limit?

Table 1. Finding limits by substituting values of x “close to a.” (Values of f (x) and g(x) rounded

to six decimals.)

¹is idea of making beer and beer approximations is actually a common approach in modern science:

if you have a complicated problem (such as predicting tomorrow’s weather, or the next three decades’ climate)

that you cannot solve directly, one thing oen tried is changing the problem a lile bit, and then making ap-

proximations with a computer. is activity is called Scientific Computation, a very active branch of modern

mathematics.


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32 III. LIMITS AND CONTINUOUS FUNCTIONS

1.5. Example: substituting numbers can suggest the wrong answer. Suppose wehad the function

g(x) = 101000x

100 000x + 1

and we want to find the limit limx→0 g(x).en substitution of some “small values of x” could lead us to believe that the limit is

1.000 . . .. But in fact, if we substitute sufficiently small values, we will see that the limitis 0 (zero)!

As you see from this example, there’s more to evaluating limits than just typing num-bers into the computer and hiing return. See also problem 12.

2. Problems

1. Guess limx→2 x10. Then try using thefirst definition of the limit to show that your

guess is right.

2. Use a calculator to guess the value of

limx→0

(1 + x

1/x.

3. Is the limit limx→0(1 + 0.693x)1/x aninteger? Give reasons for your answer, andcompare with your neighbor’s answer.

4. Simplicio computed 2−10 ≈ 0.001 whichis very close to zero. He therefore concludes

that if x is very close to 2, then x−10

is veryclose to zero, so that, according to Simplicio,

it is clearly true that

limx→2 x−10

= 0.Comment on Simplicio’s reasoning: do youagree with his answer? Do you agree withhis reasoning?

5. Use a calculator to guess

(a) limx→1

x100

1.01 + x100.

(b) limx→1

1.01 − x100x100

.

(c) limx→1

x100

1.01 − x100 .

3. e formal, authoritative, definition of limit

Our aempted definitions of the limit uses undefined and non-mathematical phraseslike “closer and closer”. In the end we don’t really know what those statements reallymean, although they are suggestive. Fortunately, there is a good definition, one thatis unambiguous and can be used to sele any dispute about the question of whetherlimx→a f (x) equals some number L or not. Here is the definition.

3.1. Definition of limx→a f (x) = L. If f (x) is a function defined for all x in some interval which contains a, except possibly at x = a, then we say that L is the limit of f (x)as x

→a, if for every ε > 0, we can find a δ > 0 (depending on ϵ) such that for all x in the

domain of f it is true that

0 < |x − a| < δ implies |f (x) − L| < ε. (10)

Why the absolute values? e quantity |x − y| is the distance between the points xand y on the number line, and one can measure how close x is to y by calculating |x − y|.e inequality |x − y| < δ says that “the distance between x and y is less than δ ,” or that“x and y are closer than δ .”


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3. THE FORMAL, AUTHORITATIVE, DEFINITION OF LIMI T 33

What are ε and δ ? e quantity ε is how close you would like f (x) to be to its limitL; the quantity δ is how close you have to choose x to a to achieve this. To prove thatlim

x→af (x) = L you must assume that someone has given you an unknown ε > 0, and

then find a positive δ for which (10) holds. e δ you find will depend on ε.

3.2. Show that limx→5(2x + 1) = 11 . We have f (x) = 2x + 1, a = 5 and L = 11,and the question we must answer is “how close should x be to 5 if want to be sure thatf (x) = 2x + 1 differs less than ε from L = 11?”

To figure this out we try to get an idea of how big |f (x) − L| is:|f (x) − L| =

(2x + 1) − 11 = |2x − 10| = 2 · |x − 5| = 2 · |x − a|.So, if 2|x − a| < ε then we have |f (x) − L| < ε, i.e.,

if |x − a| < 12ε then |f (x) − L| < ε.We can therefore choose δ = 12ε. No maer what ε > 0 we are given our δ will also bepositive, and if

|x

−5

|< δ then we can guarantee

|(2x + 1)

−11

|< ε. at shows that

limx→5 2x + 1 = 11.

3.3. e limit limx→1 x2 = 1, the triangle inequality, and the “don’t oose δ > 1”tri. is example will show you two basic tricks that are useful in many ε-δ arguments.

e problem is to show that x2 goes to 1 as x goes to 1: we have f (x) = x2, a = 1,L = 1, and again the question is, “how small should |x−1| be to guarantee |x2−1| < ε?”

We begin by estimating the difference |x2 − 1||x2 − 1| = |(x − 1)(x + 1)| = |x − 1| · |x + 1|.

How big can the two factors |x − 1| and |x + 1| be when we assume |x − 1| < δ ? Clearlythe first factor |x − 1| satisfies |x − 1| < δ because that is what we had assumed. For the The triangle inequality s

that

|a + b| ≤ |a| + |b|for any two real numbe

second factor we have

|x + 1

|=

|x

−1 + 2

| ≤ |x

−1

|+

|2

| by the triangle inequality ≤ δ + 2.It follows that if |x − 1| < δ then

|x2 − 1| ≤ (2 + δ )δ.Our goal is to show that if δ is small enough then the estimate on the right will not bemore than ε. Here is the second trick in this example: we agree that we always choose our δ so that δ ≤ 1. If we do that, then we will always have

(2 + δ )δ 0 is.


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L − ε

L + ε

L

y = f (x)

a

How close must x be to a for f (x) to end up in this range?

L − ε

L + ε

L

y = f (x)

a + δ a − δ a

For some x in this interval f (x) is not between L − ε anL + ε. Therefore the δ in this picture is too big for the giveε. You need a smaller δ .

L − ε

L + ε

L

y = f (x)

a + δ a − δ a


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We begin by estimating |f (x) − 14 | in terms of |x − 4|:

|f (x)

−1/4

|= 1x −

1

4 = 4 − x

4x = |x − 4|

|4x| =

1

|4x| |x

−4|.

As before, things would be easier if 1/|4x| were a constant. To achieve that we againagree not to take δ > 1. If we always have δ ≤ 1, then we will always have |x − 4| 0 Joe plansto mark off a length x and cut out a square of side x. Bruce asks Joe for a square with area 4 squarefoot. Joe tells Bruce that he can’t measure exactly 2 feet and the area of the square he produces willonly be approximately 4 square feet. Bruce doesn’t mind as long as the area of the square doesn’tdiffer more than 0.01 square feet from what he really asked for (namely, 4 square foot).

(a) What is the biggest error Joe can afford to make when he marks off the length x?

(b) Jen also wants square sheets, with area 4 square feet. However, she needs the error in the areato be less than 0.00001 square foot. (She’s paying.)

How accurately must Joe measure the side of the squares he’s going to cut for Jen?

2. [Group Problem] (Joe goes cubic.) Joe is offering to build cubes of side x. Airline regula-tions allow you take a cube on board provided its volume and surface area add up to less than 33(everything measured in feet). For instance, a cube with 2-foot sides has volume+area equal to23 + 6 × 22 = 32.

If you ask Joe to build a cube whose volume plus total surface area is 32 with an error of atmost ε, then what error can he afford to make when he measures the side of the cube he’s making?

3. Our definition of a derivative in (9) contains a limit. What is the function “f ” there, and whatis the variable? [ A]

Use the ε– δ definition to prove the following limits:

4. limx→1 2x − 4 = 6 [ A]

5. limx→2

x2 = 4. [ A]

6. limx→2

x2 − 7x + 3 = −7[ A]

7. limx→3 x3 = 27 [ A]

8. limx→2

x3 + 6x2 = 32.

9. limx→4

√ x = 2. [ A]

10. limx→3

√ x + 6 = 9. [ A]

11. limx→2 1 + x4 + x =

12 . [ A]

12. limx→1

2 − x4 − x =

13

.

13. limx→3

x

6 − x = 1.

14. limx→0

√ |x| = 0


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5. VARIATIONS ON THE LIMIT THEME 37

5. Variations on the limit theme

Not all limits are “for x

→ a”. Here we describe some variations on the concept of

limit.

5.1. Le and right limits. When we let “x approach a” we allow x to be larger orsmaller than a, as long as x “gets close to a”. If we explicitly want to study the behaviorof f (x) as x approaches a through values larger than a, then we write

limx↘a

f (x) or limx→a+

f (x) or limx→a+0

f (x) or limx→a,x>a

f (x).

All four notations are commonly used. Similarly, to designate the value which f (x) ap-proaches as x approaches a through values below a one writes

limx↗a

f (x) or limx→a−

f (x) or limx→a−0

f (x) or limx→a,x 0 one can find a δ > 0 such that

a < x < a + δ =⇒ |f (x) − L| < εholds for all x in the domain of f .

5.3. Definition of le-limits. Let f be a function. en

limx↘a

f (x) = L. (12)

means that for every ε > 0 one can find a δ > 0 such that

a − δ < x < a =⇒ |f (x) − L| < εholds for all x in the domain of f .

e following theorem tells you how to use one-sided limits to decide if a functionf (x) has a limit at x = a.

5.4. eorem. e two-sided limit limx→a

f (x) exists if and only if the two one-sided

limits

limx↘a

f (x), and limx↗a

f (x)

exist and have the same value.

5.5. Limits at infinity. Instead of leing x approach some finite number, one can letx become “larger and larger” and ask what happens to f (x). If there is a number L suchthat f (x) gets arbitrarily close to L if one chooses x sufficiently large, then we write

limx→∞ f (x) = L(“e limit for x going to infinity is L.”)

We have an analogous definition for what happens to f (x) as x becomes very largeand negative: we write

limx→−∞

f (x) = L

(“e limit for x going to negative infinity is L.”)Here are the precise definitions:


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Let f (x) be a function which is defined on an interval x0 < x < ∞. If there is a number L such that for every ε > 0 we can find an A such that

x > A =⇒ |f (x) − L| < ε for all x, then we say that the limit of f (x) for x → ∞ is L.

Let f (x) be a function which is defined on an interval −∞ < x < x0. If there is a number L such that for every ε > 0 we can find an A such that

x < −A =⇒ |f (x) − L| < ε for all x, then we say that the limit of f (x) for x → ∞ is L.

ese definitions are very similar to the original definition of the limit. Instead of δ which specifies how close x should be to a, we now have a number A that says how largex should be, which is a way of saying “how close x should be to infinity” (or to negativeinfinity).

A

+ε

−ε

Here A is too small, because f (x) > ε happens for some x ≥ A

A

+ε

−ε

This A is large enough, because

f (x) is in the right range for all x ≥ A

Figure 1. The limit of f (x) as x → ∞; how large must x be if you need −ε < f (x) < ε?

5.6. Example – Limit of 1/x. e larger x is, the smaller its reciprocal is, so it seemsnatural that 1/x → 0 as x → ∞. To prove that limx→∞ 1/x = 0 we apply the definitionto f (x) = 1/x, L = 0.

For a given ε > 0, we need to show that 1x − L < ε for all x > A (13)


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6. PROPERTIES OF THE LIMIT 39

provided we choose the right A.How do we choose A? A is not allowed to depend on x, but it may depend on ε.Let’s decide that we will always take A > 0, so that we only need consider positive

values of x. en (13) simplifies to1

x < ε

which is equivalent to

x > 1

ε.

is tells us how to choose A. Given any positive ε, we will simply choose

A = the larger of 0 and 1

ε

en we have | 1x − 0| = 1

x < ε for all x > A, so we have proved that limx→∞ 1/x = 0.

6. Properties of the Limit

e precise definition of the limit is not easy to use, and fortunately we don’t needto use it very oen. Instead, there are a number of properties that limits possess that willallow us to compute complicated limits without having to resort to “epsiloncy.”

e following properties also apply to the variations on the limit from §5. I.e., thefollowing statements remain true if one replaces each limit by a one-sided limit, or a limitfor x → ∞.

Limits of constants and of x. If a and c are constants, then

limx→a

c = c (P 1)

and

limx→a

x = a. (P 2)

Limits of sums, products, powers, and quotients. Let F 1 and F 2 be two givenfunctions whose limits for x → a we know,limx→a

F 1(x) = L1, limx→a

F 2(x) = L2.

en

limx→a

(F 1(x) + F 2(x)

) = L1 + L2, (P 3)

limx→a

(F 1(x) − F 2(x)

) = L1 − L2, (P 4)

limx→a

(F 1(x) · F 2(x)

) = L1 · L2. (P 5)

If limx→a F 2(x)̸ = 0, then

limx→a F 1(x)F 2(x) =

L1L2

. (P 6)

Finally, if k is a positive real number, then

limx→a

(F 1(x)

k)

= L1k. (P 7)

In other words the limit of the sum is the sum of the limits, etc. One can prove theselaws using the definition of limit in §3 but we will not do this here. However, these lawsshould seem like common sense: if, for x close to a, the quantity F 1(x) is close to L1 and


40/195


F 2(x) is close to L2, then certainly F 1(x) + F 2(x) should be close to L1 + L2. (And soforth.)

Later in this chapter we will add two more properties of limits to this list. ey arethe “Sandwich eorem” (§11) and the substitution theorem (§12).

7. Examples of limit computations

7.1. Find limx→2 x2. We have

limx→2

x2 = limx→2

x · x=(

limx→2

x) · ( lim

x→2x)

by (P 5)

= 2 · 2 = 4.Similarly,

limx→2

x3 = limx→2

x · x2

= ( limx→2 x) · ( limx→2 x2) (P 5) again= 2 · 4 = 8,

and, by (P 4)limx→2

x2 − 1 = limx→2

x2 − limx→2

1 = 4 − 1 = 3,and, by (P 4) again,

limx→2

x3 − 1 = limx→2

x3 − limx→2

1 = 8 − 1 = 7,Puing all this together, we get

limx→2

x3 − 1x2 − 1 =

23 − 122 − 1 =

8 − 14 − 1 =

7

3

because of (

P 6)

. To apply (

P 6)

we must check that the denominator (“L2

”) is not zero.Since the denominator is 3 everything is OK, and we were allowed to use (P 6).

7.2. Try the examples 1.4 and 1.5 using the limit properties. To compute limx→2(x2−2x)/(x2 − 4) we first use the limit properties to find

limx→2

x2 − 2x = 0 and limx→2

x2 − 4 = 0.to complete the computation we would like to apply the last property (P 6) about quo-tients, but this would give us

limx→2

f (x) = 0

0.

e denominator is zero, so we were not allowed to use (P 6) (and the result doesn’t meananything anyway). We have to do something else.

e function we are dealing with is a rational function, which means that it is thequotient of two polynomials. For such functions there is an algebra trick that alwaysallows you to compute the limit even if you first get 00 . e thing to do is to dividenumerator and denominator by x − 2. In our case we have

x2 − 2x = (x − 2) · x, x2 − 4 = (x − 2) · (x + 2)so that

limx→2

f (x) = limx→2

(x − 2) · x(x − 2) · (x + 2) = limx→2

x

x + 2.


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7. EXAMPLES OF LIMIT COMPUTATIONS 41

Aer this simplification we can use the properties (P ...) to compute

limx→2 f (x) =

2

2 + 2 =

1

2 .

7.3. Example – Find limx→2√

x. We can apply limit property (P 5a) with k = 1/2to see that

limx→2

√ x = lim

x→2x1/2 = [ lim

x→2x]1/2 = 21/2 =

√ 2.

7.4. Example – e derivative of √

x at x = 2. Find

limx→2

√ x − √ 2x − 2

assuming the result from the previous example (i.e., assuming that limx→2√

x =√

2.)Solution: e function is a fraction whose numerator and denominator vanish when

x = 2, so the limit is of the form “00 ”. We use an algebra trick; namely, multiplying the

numerator and denominator by√

x +√

2:

√ x − √ 2x − 2 =

(√

x − √ 2)(√ x + √ 2)(x − 2)(√ x + √ 2) =

1√ x +

√ 2

.

Now we can use the limit properties to compute

limx→2

√ x − √ 2x − 2 = limx→2

1√ x +

√ 2

= 1

2√

2=

√ 2

4 .

7.5. Limit as x → ∞

of Rational Functions. A rational function is the quotient of two polynomials:

R(x) = anx

n + · · · + a1x + a0bmxm + · · · + b1x + b0 . (14)

We have seen that

limx→∞

1

x = 0

We even proved this in example 5.6. Using this we can find the limit at ∞ for any rationalfunction R(x) as in (14). We could turn the outcome of the calculation of limx→∞ R(x)into a boxed recipe involving the degrees n and m of the numerator and denominator,and also their coefficients ai, bj , which you, the student, would then memorize, but it isbeer to remember the trick:

To find the limit as x → ∞,of some rational function you have been given, factor the highest occurring powers of x,

both from numerator and from denominator.

For example, let’s compute

limx→∞

3x2 + 3

5x2 + 7x − 39 .


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Remember the trick and factor x2 from top and boom. You get

limx→∞3x2 + 3

5x2 + 7x − 39 = limx→∞x2

x2

3 + 3/x2

5 + 7/x − 39/x2 (algebra)= lim

x→∞3 + 3/x2

5 + 7/x − 39/x2 (more algebra)

= limx→∞(3 + 3/x2)

limx→∞(5 + 7/x − 39/x2) (limit properties)

= 3

5.

At the end of this computation, we used the limit properties (P ∗) to break the limit downinto simpler pieces like limx→∞ 39/x2, which we can directly evaluate; for example, wehave

limx→∞39/x2

= limx→∞ 39 · 1x2

= limx→∞39 · limx→∞ 1x2

= 39 · 02

= 0.

e other terms are similar.

7.6. Another example with a rational function. Compute

limx→∞

2x

4x3 + 5.

We apply “the trick” again and factor x out of the numerator and x3 out of the denomi-nator. is leads to

limx→∞

2x

4x3 + 5 = lim

x→∞ x

x32

4 + 5/x3= lim

x→∞ 1

x22

4 + 5/x3

= limx→∞

1x2

·

limx→∞

2

4 + 5/x3

= 0 · 24= 0.

is example and the previous cover two of the three possible combinations of degrees nand m, namely n = m and n < m. To show the remaining case, in which the numeratorhas higher degree than the denominator, we should do yet another example, but that onewill have to wait a few pages (see § 9.3)

8. When limits fail to exist

In the last couple of examples we worried about the possibility that a limit limx→a g(x)actually might not exist. is can actually happen, and in this section we’ll see a few ex-amples of what failed limits look like. First let’s agree on what we will call a “failed limit.”

8.1. Definition. If there is no number L such that limx→a f (x) = L, then we say that the limit limx→a f (x) does not exist.


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8. WHEN LIMITS FAIL TO EXIST 43

8.2. e sign function near x = 0. e “sign function” is defined by

1

−1

y = sign(x)There are those who do not

like the notation “sign(x),”and prefer to write

g(x) = x|x| instead of g(x) = sign(x). If you

think about this formula

for a moment you’ll see

that sign(x) and x/|x| arethe same for all x̸ = 0 .

When x = 0 the quotient

x/|x| is of course notdefined.

sign(x) = −1 for x 0

Note that “the sign of zero” is defined to be zero. Our question is: does the sign functionhave a limit at x = 0? e answer is no , and here is why: since sign(x) = +1 for allpositive values of x, we see that

limx↘0 sign(x) = +1.

Similarly, since sign(x) = −1 for all negative values of x, we see thatlimx↗0 sign(x) = −1.

en by the result of 5.4, we see that because the two one-sided limits have differentvalues, the two-sided limit does not exist. (Essentially, if the two-sided limit existed, then

it would have to be equal to both +1 and −1 at the same time, which is impossible.)Conclusion: lim

x→0sign(x) does not exist.

8.3. e example of the baward sine. Contemplate the limit as x → 0 of the“backward sine,” i.e.

limx→0

sinπ

x

.

When x = 0 the function f (x) = sin(π/x) is not defined, because its definitioninvolves division by x. What happens to f (x) as x → 0? First, π/x becomes larger andlarger (“goes to infinity”) as x → 0. en, taking the sine, we see that sin(π/x) oscillatesbetween +1 and −1 infinitely oen as x → 0, since, for example, we can calculate thatf ( 24k+1 ) = +1 and f (

24k+3 ) =

−1 for each integer k. is means that f (x) gets close to

any number between −1 and +1 as x → 0, but that the function f (x) never stays closeto any particular value because it keeps oscillating up and down between +1 and −1.

y = sin πx

1 2 3

12

A

B

C

D

E

Figure 2. Graph of y = sin πx

for −3 < x < 3, x̸ = 0.

Here again, the limit limx→0 f (x) does not exist. We have arrived at this conclusionby only considering what f (x) does for small positive values of x. So the limit fails toexist in a stronger way than in the example of the sign-function. ere, even though the


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limit didn’t exist, the one-sided limits existed. In the present example we see that eventhe one-sided limits

limx↘0 sinπx and limx↗0 sinπxdo not exist.

8.4. Limit pro

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